Percentage Yield & Purity (Cambridge (CIE) IGCSE Chemistry): Revision Note

Copper(II) sulfate may be prepared by the reaction of dilute sulfuric acid with copper(II) oxide. A student prepared 1.6 g of dry copper(II) sulfate crystals.

Calculate the percentage yield if the theoretical yield is 2.0 g.

Answer

• Actual yield of copper(II) sulfate = 1.6 g

• Percentage yield of copper(II) sulfate = $\frac{1.6}{2.0}$ x 100 = 80%

• You are expected to remember the equation for percentage yield

• If you remember it incorrectly and get a percentage yield greater than 100%, then you have made an error!

• The most common error is to divide the theoretical yield by the actual yield

  • In this case, you just need to swap the numbers around in your calculation

Calculate the percentage by mass of iron in iron(III) oxide, Fe₂O₃.

Answer:

• From the Periodic Table, the relative atomic masses are:

  • Fe = 56

  • Oxygen = 16 

• The total mass of iron in iron(III) oxide is:

  • 2 x 56 = 112

• The total mass of iron(III) oxide is:

  • (2 × 56) + (3 × 16) = 160

• The equation for percentage composition is:

  • Percentage composition = $\frac{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{element}}{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{compound}}$ × 100

• So, the percentage composition of iron in iron(III) oxide is:

  • Percentage of iron = $\frac{112}{160}$ × 100 = 70%

The chemical formula of the fertiliser ammonium nitrate is NH₄NO₃. Calculate the percentage by mass of nitrogen in ammonium nitrate.

Answer:

• From the Periodic Table, the relative atomic masses are:

  • Nitrogen = 14

  • Hydrogen = 1

  • Oxygen = 16 

• Careful: There are two nitrogen atoms in ammonium nitrate

• The total mass of nitrogen in ammonium nitrate is:

  • 2 x 14 = 28

• The total mass of ammonium nitrate is:

  • (1 x 14) + (4 × 1) + (1 x 14) + (3 × 16) = 80

• The equation for percentage composition is:

  • Percentage Composition = $\frac{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{element}}{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{compound}}$ × 100

• So, the percentage composition of nitrogen in ammonium nitrate is:

• Percentage of nitrogen = $\frac{\left(2\times 14\right)}{80}$ × 100 = 35%

• Make sure you calculate the percentage composition using the total mass of the element.

  • For example, a common mistake with ammonium nitrate is doing the calculation for only one atom of nitrogen.

  • This would lose a mark in an exam

• Show ALL your working out. If you make a mistake in the calculation, you could still score a mark for your workings.

A sample of lead(II) bromide was made. It weighed 15 g.

The sample was found to be impure and only contained 13.5 g of lead(II) bromide. 

Calculate the percentage purity of the lead(II) bromide.

Answer:

• The mass of the pure substance is 13.5 g

• The total mass of the substance is 15 g

• Percentage purity = $\frac{\mathrm{mass}\mathrm{of}\mathrm{pure}\mathrm{substance}}{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{substance}}\mathrm{x}100$

• Percentage purity = $\frac{13.5}{15}\times 100$ = 90% 

• All of these calculations are to find a percentage so don't forget to multiply by 100 to convert your answer to a percentage.